Question

While landing at an airport, a pilot made an angle of depression of ${20}^o$. Average speed of the plane was $200km/hr.$ The plane reached the ground after $54$ seconds. Find the height at which the plane was when it started landing. $(\sin {20}^o=0.342)$

Answer 1

Let, $AB$ be the height of the plane when it started landing $=h$.

and $AC$ be the distance travelled by the plane during the landing.

Time taken $=54sec=\frac{54}{3600}hr$

From diagram,

$l\parallel BC\Rightarrow \angle ACB={20}^o$

$AC=Distance=Speed\times time=200\times \frac{54}{3600}km$

$=3km$

From $\Delta ABC$

$\sin {20}^o=\frac{AB}{AC}$

$\Rightarrow 0.342=\frac{h}{3}$

$\Rightarrow h=3\times 0.342$

$\Rightarrow h=1.026$

Hence, the height of the plane was $1.026km$